tracediagramreview.tex
\subsection{Gluing Lemmas}\label{s:gluinglemmas}

This section introduces some additional notation that will be used to generate the recurrence formulas in the later sections.

Finally, spin networks satisfy certain ``recoupling'' identities. These are usually written in terms of the $6j$-symbols described next, but we will introduce notation useful in simplifying the specific recurrence relations we obtain later in the paper.

\begin{definition}[$6j$-symbols, Definition 5.4 in \cite{LP}]\label{d:6jsymbol}
    The \emph{$6j$-symbols} are the coefficients in the following change-of-basis equation:
    $$
    \tikz[trivalent]{
        \coordinate(vxa)at(0,.35)
            edge[]node[leftlabel,pos=1]{$d$}(0,-.15)edge[bend left]node[leftlabel,pos=1]{$a$}(-.6,1.15);
        \coordinate(vxb)at(.3,.75)
            edge[bend left]node[basiclabel,pos=.7,right=2pt]{$e$}(vxa)
            edge[bend left]node[leftlabel,pos=1]{$b$}(0,1.15)edge[bend right]node[rightlabel,pos=1]{$c$}(.6,1.15);}
    = \sum_{f\in\lceil a,b\rfloor\cap\lceil c,d\rfloor} \sixj abdcef
    \tikz[trivalent]{
        \coordinate(vxa)at(0,.35)
            edge[]node[leftlabel,pos=1]{$d$}(0,-.15)edge[bend right]node[rightlabel,pos=1]{$c$}(.6,1.15);
        \coordinate(vxb)at(-.3,.75)
            edge[bend right]node[basiclabel,pos=.7,left=3pt]{$f$}(vxa)
            edge[bend right]node[rightlabel,pos=1]{$b$}(0,1.15)edge[bend left]node[leftlabel,pos=1]{$a$}(-.6,1.15);}
    .$$
\end{definition}
Note that this definition differs slightly from those found in the literature \cite{CFS95,Kau91}.

Together with the fusion identity, this provides an identity for gluing a strand ``across'' a vertex.
\begin{proposition}[Strand-Vertex Gluing]\label{l:vertexglue}
    $$
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.5)
        edge[]node[rightlabel,pos=1]{$a$}(.5,-.4)
        edge[bend left]node[rightlabel,pos=1]{$c$}(0,1.1)
        edge[bend right]node[rightlabel,pos=1]{$b$}(1,1.1);
        \draw[blue](.3,-.4)to(.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{$d$};}
    = \sum_{e\in\iadm{a,d}} \mff_e(a,d)
    \tikz[trivalent]{
        \coordinate(vxa)at(0,.35)
            edge[blue,bend left]node[leftlabel,pos=1]{$d$}(-.6,1.15);
        \coordinate(vxc)at(0,-.15)
            edge[]node[leftlabel]{$e$}(vxa)
            edge[blue,bend right]node[leftlabel,pos=1]{$d$}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{$a$}(.3,-.55);
        \coordinate(vxb)at(.3,.75)
            edge[bend left]node[basiclabel,pos=.7,right=2pt]{$a$}(vxa)
            edge[bend left]node[leftlabel,pos=1]{$c$}(0,1.15)edge[bend right]node[rightlabel,pos=1]{$b$}(.6,1.15);}
    = \sum_{\substack{e\in\iadm{a,d}\\ f\in\iadm{c,d}\cap\iadm{b,e}}}
    \mff_e(a,d)\sixj dcebaf
    \tikz[trivalent]{
        \coordinate(vxa)at(0,.35)
            edge[bend right]node[rightlabel,pos=1]{$b$}(.6,1.15);
        \coordinate(vxc)at(0,-.15)
            edge[]node[leftlabel]{$e$}(vxa)
            edge[blue,bend right]node[leftlabel,pos=1]{$d$}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{$a$}(.3,-.55);
        \coordinate(vxb)at(-.3,.75)
            edge[bend right]node[basiclabel,pos=.7,left=3pt]{$f$}(vxa)
            edge[bend right]node[rightlabel,pos=1]{$c$}(0,1.15)edge[blue,bend left]node[leftlabel,pos=1]{$d$}(-.6,1.15);}
    .$$
\begin{proof}
    The first step applies the fusion identity (Proposition \ref{p:bubblefusion}) to the lower part of the diagram, introducing a summation over $e$. The second step uses the change-of-basis formula (Definition \ref{d:6jsymbol}) on the upper half of the diagram.
\end{proof}
\end{proposition}

We now introduce specialized notation for the case of the above lemma with $d=1$, since it will simplify the expression of the product in Theorem \ref{t:simplerecurrence}.

\begin{definition}\label{d:fusioncoeff}
    Given an admissible triple $\{a,b,c\}$, $a'\in\iadm{1,a}$, and $c'\in\iadm{1,c}$, define the \emph{fusion coefficient} $\mathfrak{F}$ and the \emph{normalized fusion coefficient} $\hat{\mathfrak{F}}$ by
    \begin{align*}
        \Fus ba{a'}c{c'} \equiv \mfs_{a'}(1,a) \mff_{a'}(1,a) \sixjt 1c{a'}ba{c'}.\\
        \nFus ba{a'}c{c'} \equiv \mfs_{a'}(1,a) \sqrt{\tfrac{\mff_{a'}(1,a)}{\mff_{c'}(1,c)}} \sixjt 1c{a'}ba{c'}.
    \end{align*}
\end{definition}

It is immediate that
    \begin{equation}\label{eq:fusnormalization}
        \Fus ba{a'}c{c'}=\sqrt{\mff_{a'}(1,a)\mff_{c'}(1,c)} \nFus ba{a'}c{c'}.
    \end{equation}
Also, $a'\in\iadm{1,a}$ is equivalent to requiring $a'=a\pm1$, so given $\{a,b,c\}$ there are four choices for $\{a',c'\}$. Table \ref{t:fusioncoeff} shows the values of the $6j$-symbol, the fusion coefficient, and the reduced coefficient in each of the four cases. Observe that both the fusion coefficient and its normalized version are invariant under the interchange of $a$ and $c$:
    $$\Fus ba{a'}c{c'}=\Fus bc{c'}a{a'} \quad\text{and}\quad \nFus ba{a'}c{c'}=\nFus bc{c'}a{a'}.$$
These values are taken from Corollary 5.5 in \cite{LP}, together with the fact that the additional fusion constant is either $\mff_{a+1}(1,a)=1$ or $\mff_{a-1}(1,a)=\frac{a}{a+1}$. This table corrects sign errors in the $6j$-symbol formulas provided on pp. 79-80 of \cite{Pet06}.

    \def\spacer{\tikz{\draw[white,transparent](0,-.2)--(0,1.2);}}
    \def\spacerb{\tikz{\draw[white,transparent](0,-.4)--(0,1.4);}}
    \begin{center}
    \begin{table}
    \begin{tabular}{|c|ccc|}
    \hline
    \spacer
    $\{a',c'\}$ & $\sixjt 1c{a'}ba{c'}$ & $\Fus ba{a'}c{c'}$ & $\nFus ba{a'}c{c'}$ \\\hline
    \spacer
    $\{a+1,c+1\}$ & $1$ & $1$ & $1$\\
    \spacer
    $\{a-1,c+1\}$ & $\tfrac{\mfe_c(a,b)}{a}$ & $-\tfrac{\mfe_c(a,b)}{a+1}$
        & $-\tfrac{\mfe_c(a,b)}{\sqrt{a(a+1)}}$ \\
    \spacer
    $\{a+1,c-1\}$ & $-\tfrac{\mfe_a(c,b)}{c+1}$ & $-\tfrac{\mfe_a(c,b)}{c+1}$
        & $-\tfrac{\mfe_a(c,b)}{\sqrt{c(c+1)}}$ \\
    \spacerb
    $\{a-1,c-1\}$ & $\frac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{a(c+1)}$ & $-\tfrac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{(a+1)(c+1)}$
        & $-\tfrac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{\sqrt{a(a+1)c(c+1)}}$ \\\hline
    \end{tabular}
    \caption{The four basic fusion coefficients, with corresponding $6j$-symbols and normalized values.}\label{t:fusioncoeff}
    \end{table}
    \end{center}

With a different orientation, Proposition \ref{l:vertexglue} becomes
\begin{proposition}\label{l:gluevertex-alt}
    \begin{equation}\label{eq:glueonesum2}
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{$b$}(.5,-.1)
            edge[bend left]node[leftlabel,pos=1]{$c$}(0,1.1)
            edge[bend right]node[rightlabel,pos=1]{$a$}(1,1.1);
        \draw[blue](.2,1.1)to[out=-90,in=-90,looseness=2](.8,1.1)node[leftlabel]{$1$};}
    = \sum_{\substack{a'\in\iadm{1,a}\\ c'\in\iadm{1,c}\cap\iadm{b,a'}}}
        \Fus ba{a'}c{c'}
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{$b$}(.5,-.1);
        \coordinate(vxb)at(.1,.8)edge[bend right]node[leftlabel]{$c'\:$}(vxa)
            edge[bend left]node[leftlabel,pos=1]{$c$}(-.2,1.2)
            edge[blue,bend right]node[leftlabel,pos=1]{$1$}(.4,1.2);
        \coordinate(vxc)at(.9,.8)edge[bend left]node[rightlabel]{$a'$}(vxa)
            edge[blue,bend left]node[rightlabel,pos=1]{$1$}(.6,1.2)
            edge[bend right]node[rightlabel,pos=1]{$a$}(1.2,1.2);}.
    \end{equation}
\begin{proof}
    In terms of the fusion coefficient, Proposition \ref{l:vertexglue} is
    \begin{equation}\label{eq:gluevertexleft}
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.5)
        edge[]node[rightlabel,pos=1]{$a$}(.5,-.4)
        edge[bend left]node[rightlabel,pos=1]{$c$}(0,1.1)
        edge[bend right]node[rightlabel,pos=1]{$b$}(1,1.1);
        \draw[blue](.3,-.4)to(.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{$1$};}
    = \sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}}
    \mfs_{a'}(a,1)\Fus ba{a'}c{c'}
    \tikz[trivalent]{
        \coordinate(vxa)at(0,.35)
            edge[bend right]node[rightlabel,pos=1]{$b$}(.6,1.15);
        \coordinate(vxc)at(0,-.15)
            edge[]node[leftlabel]{$a'$}(vxa)
            edge[bend right]node[blue,leftlabel,pos=1]{$1$}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{$a$}(.3,-.55);
        \coordinate(vxb)at(-.3,.75)
            edge[bend right]node[basiclabel,pos=.7,left=3pt]{$c'$}(vxa)
            edge[bend right]node[rightlabel,pos=1]{$c$}(0,1.15)edge[blue,bend left]node[leftlabel,pos=1]{$1$}(-.6,1.15);}.
    \end{equation}
    Reflect this relation vertically using Proposition \ref{p:spinnetreflection}, and extend the strands labeled by $1$ and $c$ on both sides of the equation to obtain:
        \begin{equation}\label{eq:extendstrands}
        \tikz[trivalent]{
            \coordinate(vertex)at(.5,.5)edge[]node[rightlabel,pos=1]{$a$}(.5,1)
                edge[bend right](.1,.2)edge[bend left]node[rightlabel,pos=1]{$b$}(1,0);
            \draw[blue](-.1,.2)to[wavyup](.3,1);
            \draw[red](-.1,.2)to[out=-90,in=-90](-.5,.2)to[wavyup](-.4,1)node[rightlabel]{$1$}
                (.1,.2)to[out=-90,in=-90,looseness=1.5](-.7,.2)to[wavyup](-.6,1)node[leftlabel]{$c$};
        }=\sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}}
            \mfs_{a'}(a,1)\Fus ba{a'}c{c'}
        \tikz[trivalent]{
            \coordinate(vertexa)at(.2,.2)edge[blue,bend right](-.1,0)edge[bend left](.3,0);
            \coordinate(vertexb)at(.5,.5)edge[bend right]node[leftlabel,pos=.3]{$c'$}(vertexa)
                edge[bend left]node[rightlabel,pos=1]{$b$}(1.1,-.2);
            \coordinate(vertexc)at(.5,.8)edge[]node[rightlabel,pos=.3]{$a'$}(vertexb)
                edge[blue,bend left]node[leftlabel,pos=1]{$1$}(.3,1.2)
                edge[bend right]node[rightlabel,pos=1]{$a$}(.7,1.2);
            \draw[red](-.1,0)to[out=-90,in=-90](-.5,0)to[wavyup](-.6,1.2)node[rightlabel]{$1$}
                (.3,0)to[out=-90,in=-90,looseness=1.5](-.7,0)to[wavyup](-.8,1.2)node[leftlabel]{$c$};
        }.
        \end{equation}
    By Proposition \ref{p:spinnetsignstrong}, the relation can be straightened, with the introduction of signs $S_1$ and $S_2$, to:
        \begin{equation}\label{eq:straightenstrands}
        S_1
        \tikz[trivalent]{
            \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{$b$}(.5,-.1)
                edge[bend left]node[leftlabel,pos=1]{$c$}(0,1.1)
                edge[bend right]node[rightlabel,pos=1]{$a$}(1,1.1);
            \draw[blue](.2,1.1)to[out=-90,in=-90,looseness=2](.8,1.1)node[leftlabel]{$1$};}
        = \sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}}
            S_2\:
            \mfs_{a'}(a,1)\Fus ba{a'}c{c'}
        \tikz[trivalent]{
            \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{$b$}(.5,-.1);
            \coordinate(vxb)at(.1,.8)edge[bend right]node[leftlabel,pos=.5]{$c'\:$}(vxa)
                edge[bend left]node[leftlabel,pos=1]{$c$}(-.2,1.2)
                edge[blue,bend right]node[leftlabel,pos=1]{$1$}(.4,1.2);
            \coordinate(vxc)at(.9,.8)edge[bend left]node[rightlabel,pos=.5]{$a'$}(vxa)
                edge[blue,bend left]node[rightlabel,pos=1]{$1$}(.6,1.2)
                edge[bend right]node[rightlabel,pos=1]{$a$}(1.2,1.2);}.
        \end{equation}

    To calculate $S_1S_2$, we must count the number of kinks in each diagram. No strands in \eqref{eq:straightenstrands} are kinked. The strands between $b$ and $c$ on the lefthand side of \eqref{eq:extendstrands} are kinked, giving a factor $S_1=\mfs_a(b,c)$. On the righthand side, strands between $b$ and either $1$ or $c$ are kinked, and so are those between $1$ and $c$, producing a factor $S_2=\mfs_{a'}(b,c')\mfs_{c'}(1,c)$. Hence
        $$S_1S_2=\mfs_{c'}(1,c)\mfs_{a'}(b,c')\mfs_a(b,c)=(-1)^{\frac12(1-(a'-a))}=\mfs_{a'}(1,a).$$
    Therefore, the coefficient is $\left(\mfs_{a'}(1,a)\right)^2\Fus ba{a'}c{c'}=\Fus ba{a'}c{c'}$.
\end{proof}
\end{proposition}

The reflective symmetries guaranteed by Proposition \ref{p:spinnetreflection} imply that corresponding formulas for the products
    \begin{equation}\label{eq:gluevertexlist}
%    \tikz[trivalent]{
%        \coordinate(vxa)at(.5,.5)
%            edge[]node[rightlabel,pos=1]{$a$}(.5,-.1)
%            edge[bend left]node[rightlabel,pos=1]{$c$}(0,1.1)
%            edge[bend right]node[rightlabel,pos=1]{$b$}(1,1.1);
%        \draw[blue](.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{$1$};}
%    \quad
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.5)
            edge[]node[leftlabel,pos=1]{$a$}(.5,-.1)
            edge[bend left]node[leftlabel,pos=1]{$b$}(0,1.1)
            edge[bend right]node[leftlabel,pos=1]{$c$}(1,1.1);
        \draw[blue](.7,-.1)to[wavyup](1.2,1.1)node[rightlabel]{$1$};}
    \quad
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.5)
            edge[]node[rightlabel,pos=1]{$a$}(.5,1.1)
            edge[bend left]node[rightlabel,pos=1]{$b$}(1,-.1)
            edge[bend right]node[rightlabel,pos=1]{$c$}(0,-.1);
        \draw[blue](-.2,-.1)to[wavyup](.3,1.1)node[leftlabel]{$1$};}
    \quad
    \tikz[trivalent]{
        \coordinate(vxa)at(.5,.5)
            edge[]node[leftlabel,pos=1]{$a$}(.5,1.1)
            edge[bend left]node[leftlabel,pos=1]{$c$}(1,-.1)
            edge[bend right]node[leftlabel,pos=1]{$b$}(0,-.1);
        \draw[blue](1.2,-.1)to[wavyup](.7,1.1)node[rightlabel]{$1$};}
    \end{equation}
can be generated by reflecting \eqref{eq:gluevertexleft}. In particular, the terms in these summations have the same coefficients.

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